Estimations for the size of the biggest entry in an inverse Matrix

Question

If you got a Matrix $A$. Is there a estimation how big the largest element in the inverse of the matrix is?

If it helps the matrix is unimodular, that means all entries are integer and the determinant is +-1. The inverse is also unimodular.

Answer 1

If the matrix is unimodular ( the determinant is $1$) then the inverse is the adjunct matrix $\text{adj} A$ ( in general it's $\frac{1}{\det A} \cdot \text{adj} A$). The matrix $\text{adj} A$ has as $(i,j)$ entry the algebraic complement of $a_{ij}$ in the expansion of the determinant $\det A$, that is, it's $(-1)^{i+j} M_{ji}$ where the minor $M_{ji}$ ( note the flipping of $i$, $j$ ) is the $n-1\times n-1 $ determinant obtained from $\det A$ by eliminating the row $j$ and column $i$. An estimate for $M_{ji}$ is given by the Hadamard's inequality. It is not clear to me right now that you cannot give a finer estimate that takes into account that the determinant of $A$ is $1$. This appears to be an interesting problem.

$\bf{Added}$ It seems that this exponent $n-1$ cannot be avoided, there is no linear estimate. For take a diagonal unimodular matrix, with diagonal $(\lambda_1, \ldots, \lambda_n)$. Then $\frac{1}{\lambda_i} = \prod_{j \ne i} \lambda_j$. One could use the singular value decomposition and reach a similar estimate.

Answer 2

Since the max norm is dependent only on the largest element, you can use perturbation theorem

http://en.wikipedia.org/wiki/Sherman%E2%80%93Morrison_formula

Choose $\bf v,u$ to be vectors, with one $\epsilon$ each and rest 0.

Numerator : $\epsilon^2 {A^{-1}_{:,j} A^{-1}_{i,:}}$ Denominator : $1 - \epsilon^2 A_{ij}^{-1}$

For any (i,j) we can design $\epsilon$ so that $\epsilon^2A_{ij}^{-1}$ is arbitrary close to $1$. Then $\epsilon^2$ will be arbitrarily close to $\frac{1}{A_{ij}^{-1}}$. But $A_{ij}^{-1}$ occurs as a square in the numerator, so one factor will survive at position (i,j), while the denominator is arbitrarily close to 0.

Answer 3

Just for illustration how bad it can be: consider an $n\times n$ lower triangular matrix $A$ with $1$ on the diagonal and $-1$ in the strictly lower triangular part. The matrix is clearly unimodular. However, the entry $(n,1)$ of $A^{-1}$ is equal to $2^{n-2}$ (so it grows very fast with $n$ even though $A$ is full of quite "innocent" numbers).